Let be a finite extension of with uniformizer , , (resp. ) the upper (resp. lower) triangular Borel, the diagonal maximal torus. Fix another finite extension , and let be a continuous character, identified with a tuple of characters in the obvious way. Suppose for all we have the inequality , with equality for . Does the locally analytic induction

admit a -invariant norm?

The answer in this generality is definitely “no”. Here’s a corrected version of this speculation.

Definition. A character as above is weakly unitary if

for all , with equality for . A character is strongly unitary if every character appearing in is weakly unitary.

Here denotes Emerton’s locally analytic Jacquet module; the definition of strong unitarity can be unwound (cf. Remark 5.1.8 in Emerton’s second Jacquet paper) into something which doesn’t explicitly mention Jacquet modules. Strong unitary is a necessary condition for to admit a -invariant norm, and now we may conjecture the converse.

Conjecture. The representation admits a -invariant norm if and only if is strongly unitary.

This is closely related to the Breuil-Schneider conjecture. Indeed, if is locally algebraic with -dominant algebraic part, then a theorem of Hu (Yongquan Hu, Normes invariantes et existence de filtrations admissibles, Crelle 634) shows that is strongly unitary if and only if is the parameter of a de Rham and trianguline (“semistabelline”) representation with distinct Hodge-Tate weights, in which case the Breuil-Schneider conjecture predicts that a certain locally algebraic representation , which coincides with the space of locally algebraic vectors in this case, admits a -invariant norm. So the above conjecture essentially implies the Breuil-Schneider conjecture for semistabelline representations.

Fix a prime , and let be a perfectoid Tate-Huber pair. Precisely, is a topological ring satisfying the following conditions:

is complete, and the subring of power-bounded elements is bounded,

contains a topologically nilpotent unit (a pseudouniformizer),

We may choose in 2. so that in and so that the th power map induces an isomorphism .

Conditions 1 and 2 here simply assert that is a “complete uniform Tate ring”, and condition 3 is the important perfectoid condition. This is Fontaine’s generalization of Scholze’s original definition, and in particular we need not assume that contains a perfectoid field. Note also that when has characteristic , is vacuously true for any .

For many reasons, one would like to define the Fontaine ring . When has characteristic zero, there is a canonical definition: take the divided power envelope of the canonical surjection (or equivalently, of the surjection ), p-adically complete, and then invert .

Now suppose is of characteristic , so . Here one must make an auxiliary choice in defining – in order to get something interesting, we need an auxiliary surjection where is a nonperfect characteristic ring. Choosing as above does the trick: we may define by applying the same recipe to the surjection . This certainly depends on the choice of . However, I claim:

Proposition. For any reduced rational , the -vector space is functorially independent of , and the assignment defines a (pre)sheaf of -vector spaces with Frobenius action on the category of perfectoid spaces in characteristic .

The idea here is very simple: if in , then there is a natural inclusion . (More generally, given a commutative diagram of homomorphisms of characteristic rings

with perfect, there is a ring map , with the evident meaning. In general this isn’t an injection, but it is in our case, i.e. with and .) In particular, given any large enough that , we have

so any factors through the inclusion , and we’re basically done. The assignment is a little irritating since some choice of uniformizer is still present, but we can get something totally canonical by defining the inverse limit

where denotes the set of ‘s for with the partial ordering given by divisibility in (since any two ‘s divide a common third , this is clearly a cofiltered limit). In fact we can throw away the eigenspace condition first and define the ring

which is totally canonical and functorial in characteristic perfectoid Tate-Huber pairs. When this recovers the usual .

It is a remarkable fact that for , the functor is representable by a perfectoid space, and in fact by the universal cover of a suitable p-divisible group. In particular,

by perfectness it’s enough to give an element of (), which is to say any element whatsoever (), and then this maps to . For , it seems likely that is no longer representable by any kind of adic space. This is actually an example of a diamond – it is “proetale under a perfectoid space” in a suitable sense. (This probably isn’t written down anywhere; JW kindly explained to me why it’s a diamond in the case , and the construction he provided pretty clearly generalizes.)